Let $f$ be a function defined by :
$f(x) = 0$ if $x<1$
$f(x) = (\alpha-1)x^{-\lambda}$ if $x\ge1$
1)Determine $\alpha$ and $\lambda$ such that $f(x)$ is a probability density function of a random variable with expected value equal to 2.
2)Determine the variance of $X$
3)Based on chebyshev's inequality find a real $a$ such that $P(|X-2|>a)<0.01$.
I've solved the first 2 question and got :
$\alpha = \lambda = 3 $ (Correct me if I'm wrong)
$Var(x) = \infty$ (Not so sure about this)
For the third question :
Chebyshev's inequality : $P((X-E[X])>kσ)<\frac{1}{k^2}$
Comparing this with the equation given in the 3rd question , wouldn't this imply that $a = kσ$ with $\frac{1}{k^2}=0.01$?
However due to the fact that the i got $\infty$ as the variance , the standard deviation $σ$ would also be infinity and hence a would be $\infty$.
I'm confused about this and i have no idea what I'm doing wrong.
I'd be glad if anyone can help me.(Let me know if you need me to post the calculations for 1) and 2))
Thanks in advance.